Lipschitz continuity and $l^1$ norm

Question

Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be a function defind as follows: $$ f(x) = \begin{cases} x, & \|x\| \leq 1 \\ x/ \|x\|, & \|x\| > 1 \end{cases} $$ where $\| \cdot \|$ is the $l^1$ norm.

Then it can be shown, that $\|f(x) - f(y)\| \leq 2\|x-y\|$. I need to prove, that $2$ cannot be improved. How can I do that?

Answer 1

Consider for $0 < \varepsilon$ the points $x = (1,0)$ and $y = (1, \varepsilon)$. Then $\lVert x-y\rVert = \varepsilon$, and we have $f(x) = x$, $f(y) = \frac{1}{1+\varepsilon}y$, so

$$\lVert f(x) - f(y)\rVert = \biggl\lvert 1 - \frac{1}{1+\varepsilon}\biggr\rvert + \frac{\varepsilon}{1+\varepsilon} = \frac{2\varepsilon}{1+\varepsilon} = \frac{2}{1+\varepsilon}\lVert x-y\rVert.$$